Ignatius and the Princess III(拆分数)
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Ignatius and the Princess III
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 4 Accepted Submission(s) : 4
"The second problem is, given an positive integer N, we define an equation like this:
N=a[1]+a[2]+a[3]+...+a[m];
a[i]>0,1<=m<=N;
My question is how many different equations you can find for a given N.
For example, assume N is 4, we can find:
4 = 4;
4 = 3 + 1;
4 = 2 + 2;
4 = 2 + 1 + 1;
4 = 1 + 1 + 1 + 1;
so the result is 5 when N is 4. Note that "4 = 3 + 1" and "4 = 1 + 3" is the same in this problem. Now, you do it!"
41020
542627
经典题,递推递归。
首先,我们引进一个小小概念来方便描述吧,record[n][m]是把自然数划分成所有元素不大于m的分法,例如:
当n=4,m=1时,要求所有的元素都比m小,所以划分法只有1种:{1,1,1,1};
当n=4,m=2时,。。。。。。。。。。。。。。。。只有3种{1,1,1,1},{2,1,1},{2,2};
当n=4,m=3时,。。。。。。。。。。。。。。。。只有4种{1,1,1,1},{2,1,1},{2,2},{3,1};
当n=4,m=5时,。。。。。。。。。。。。。。。。只有5种{1,1,1,1},{2,1,1},{2,2},{3,1},{4};
从上面我们可以发现:当n==1||m==1时,只有一种分法;
当n<m时,由于分法不可能出现负数,所以record[n][m]=record[n][n];
当n==m时,那么就得分析是否要分出m这一个数,如果要分那就只有一种{m},要是不分,那就是把n分成不大于m-1的若干份;即record[n][n]=1+record[n][n-1];
当n>m时,那么就得分析是否要分出m这一个数,如果要分那就{{m},{x1,x2,x3..}}时n-m的分法record[n-m][m],要是不分,那就是把n分成不大于m-1的若干份;即record[n][m]=record[n-m][m]+record[n][m-1];
那么其递归式:
record[n][m]= 1 (n==1||m==1)
record[n][n] (n<m)
1+record[n][m-1] (n==m)
record[n-m][m]+record[n][m-1] (N>m)
#include<cstdio>using namespace std;const int M=125;int main(){ int i,j,n,a[M][M]; for(i=1;i<M;i++) a[i][1]=a[1][i]=1; for(i=2;i<M;i++) for(j=2;j<M;j++) { if(i<j) a[i][j]=a[i][i]; if(i==j) a[i][j]=a[i][j-1]+1; if(i>j) a[i][j]=a[i-j][j]+a[i][j-1]; } while(~scanf("%d",&n)) printf("%d\n",a[n][n]); return 0;}
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